If log _{4} x+log _{8} x=5, then x is equal t | vedclass

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(D) Given the equation: $\log _{4} x+\log _{8} x=5$Using the change of base formula $\log _{b} a = \frac{\log a}{\log b}$,we get:$\frac{\log x}{\log 4

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$64$

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(D) Given the equation: $\log _{4} x+\log _{8} x=5$Using the change of base formula $\log _{b} a = \frac{\log a}{\log b}$,we get:$\frac{\log x}{\log 4} + \frac{\log x}{\log 8} = 5$Since $4 = 2^2$ and $8 = 2^3$,we can write:$\frac{\log x}{2 \log 2} + \frac{\log x}{3 \log 2} = 5$Taking $\frac{\log x}{\log 2}$ as a common factor:$\frac{\log x}{\log 2} (\frac{1}{2} + \frac{1}{3}) = 5$$\frac{\log x}{\log 2} (\frac{3+2}{6}) = 5$$\frac{\log x}{\log 2} (\frac{5}{6}) = 5$Multiplying both sides by $\frac{6}{5}$:$\frac{\log x}{\log 2} = 5 	imes \frac{6}{5} = 6$Using the property $\log _{b} a = \frac{\log a}{\log b}$,we have $\log _{2} x = 6$Converting to exponential form:$x = 2^6 = 64$

If $\log _{4} x+\log _{8} x=5,$ then $x$ is equal to

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